Results 1 to 2 of 2

Thread: help understanding this notation

  1. #1
    Junior Member
    Joined
    Jan 2008
    From
    Waipahu, HI
    Posts
    37

    help understanding this notation

    My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

    He wants us to use $\displaystyle A=B=\Re$, and $\displaystyle f:A\longrightarrow B$ to be $\displaystyle f(x)=e^x$ and $\displaystyle g:B\longrightarrow A$ to be $\displaystyle g(x)=\arctan x$.

    There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

    Define $\displaystyle A_0=A$ and $\displaystyle B_0=B; A_1=f(A_0)$ and $\displaystyle B_1=g(B_0)$. Then for each $\displaystyle n\in N$ define:
    $\displaystyle A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),$
    $\displaystyle B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).$

    How do I put $\displaystyle A=B=\Re$ into this? I want to say that they're all just $\displaystyle \Re$ but I'm not sure. Help!

    Thank you!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by sfitz View Post
    My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

    He wants us to use $\displaystyle A=B=\Re$, and $\displaystyle f:A\longrightarrow B$ to be $\displaystyle f(x)=e^x$ and $\displaystyle g:B\longrightarrow A$ to be $\displaystyle g(x)=\arctan x$.

    There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

    Define $\displaystyle A_0=A$ and $\displaystyle B_0=B; A_1=f(A_0)$ and $\displaystyle B_1=g(B_0)$. Then for each $\displaystyle n\in N$ define:
    $\displaystyle A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),$
    $\displaystyle B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).$

    How do I put $\displaystyle A=B=\Re$ into this? I want to say that they're all just $\displaystyle \Re$ but I'm not sure. Help!

    Thank you!!

    That is what it says, you then need to observe that $\displaystyle A_1=f(A)=f(\Re)=(0,\infty)$, and $\displaystyle B_1=g(B)=g(\Re)=(-\pi/2,\pi/2)$, and away you go.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Understanding Matrix Suffix Notation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 9th 2011, 08:25 AM
  2. Understanding Function Notation?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 3rd 2011, 10:18 AM
  3. Replies: 2
    Last Post: Jan 23rd 2011, 06:49 PM
  4. Replies: 3
    Last Post: Sep 5th 2010, 11:55 PM
  5. Need help understanding summation notation problem...
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Nov 23rd 2006, 03:51 AM

Search Tags


/mathhelpforum @mathhelpforum